Problem: $f\,^{\prime}(x)=-\dfrac{6}{x^4}$ and $f(1)=-4$. $f(-1) = $
Finding $f(x)$ We have $f'(x)=-\dfrac{6}{x^4}$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (-\dfrac{6}{x^4})\,dx \\\\ & = {2x^{-3}} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(1)=-4$. Here's what we get when we plug in $1$ : $\begin{aligned}f(1)&={2(1)^{-3}} {+ C}\\\\ &={2} {+ C} \end{aligned}$ We are given that this must equal $-4$ : $-4 = {2} {+ C}$ Solving the equation gives us ${C=-6}$. Finding $f(-1)$ Now, we have that $f(x)= {2x^{-3}} {-6}$. Let's find $f(-1)$ by plugging in $-1$ : $\begin{aligned}f(-1)&=2(-1)^{-3}-6\\\\ &=-8 \end{aligned}$ The answer $f(-1) = -8$